Problem: $ 9^{-\frac{1}{2}}$
Answer: $= \left(\dfrac{1}{9}\right)^{\frac{1}{2}}$ Figure out what goes in the blank: $\Big(? \Big)^{2}=\dfrac{1}{9}$ Figure out what goes in the blank: $\Big({\dfrac{1}{3}}\Big)^{2}=\dfrac{1}{9}$ So $9^{-\frac{1}{2}}=\left(\dfrac{1}{9}\right)^{\frac{1}{2}}=\dfrac{1}{3}$